A body of mass m1 moving with an unknown velocity of v_1\mathop i\limits^ \wedge , undergoes a collinear collision with a body of mass m2 moving with a velocity v_2\mathop i\limits^ \wedge . After collision, m1 and m2 move with velocities of v_3\mathop i\limits^ \wedge and v_4\mathop i\limits^ \wedge , respectively. If m2 = 0.5 m1 and v3 = 0.5 v1, then v1 is :-

Center of Mass PYQ — JEE Main 2019 · 8 Apr · Shift 2 · Q56 with Solution

View written solutionFree

Correct answer: C

Since the collision is collinear, momentum along the line is conserved:

$m_1 v_1 + m_2 v_2 = m_1 v_3 + m_2 v_4$

We are given:

$m_2 = 0.5m_1 = \frac{m_1}{2}$

and

$v_3 = 0.5v_1 = \frac{v_1}{2}$

Substitute these into momentum conservation:

$m_1 v_1 + \frac{m_1}{2} v_2 = m_1\left(\frac{v_1}{2}\right) + \frac{m_1}{2}v_4$

$v_1 + \frac{v_2}{2} = \frac{v_1}{2} + \frac{v_4}{2}$

Bring $\frac{v_1}{2}$ to the left:

$v_1 - \frac{v_1}{2} + \frac{v_2}{2} = \frac{v_4}{2}$

$\frac{v_1}{2} + \frac{v_2}{2} = \frac{v_4}{2}$

Multiply by 2:

$v_1 + v_2 = v_4$

Hence,

$v_1 = v_4 - v_2$

This corresponds to:

Option C: $v_4 - v_2$